# What is the ionization energy of water in kJ/mol? Is it endothermic or exothermic reaction?

Jul 29, 2017

The ionization of water is the energy associated with the process....

#### Explanation:

${H}_{2} O \left(g\right) + \Delta \rightarrow {H}_{2} {O}^{+} \left(g\right) + {e}^{-}$

Now certainly this process is endothermic......

The autoprotolysis of water is represented by the equation......

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

, and again given, ${K}_{w} = {10}^{-} 14$, this is an endothermic process. We would expect the process to be endothermic given that we break strong $H - O$ bonds.........

Jul 29, 2017

As usual, I assume gas phase because it's ionization energy and we conventionally define ionization energy for gas phase particles...

${\text{H"_2"O"(g) + Delta -> "H"_2"O}}^{+} \left(g\right) + {e}^{-}$

Fortunately, this process is well-known, and its enthalpy is:

$\textcolor{b l u e}{\Delta {H}_{I E} = \text{12.621 eV}}$

(It is interestingly easier to ionize water than $\text{N}$ atom, whose ionization energy is $\text{14.53 eV}$.)

I'll leave it as an exercise for you to convert to $\text{kJ/mol}$. Use:

• $1.602 \times {10}^{- 19} \text{J" = "1 eV}$
• $\text{1000 J}$ $=$ $\text{1 kJ}$
• $\text{1 mol e"^(-) = 6.022 xx 10^(23) "electrons}$

and you should get approximately $\underline{\text{1217.61 kJ/mol}}$. Is positive enthalpy endothermic or exothermic? Do you think it requires energy input to remove an electron? (Try any similar exercise, like removing stems from cherries, caps from pens, etc., then decide.)

See the following MO diagram: The ionization of water involves removal of the electron from its nonbonding $2 {p}_{y}$, or $1 {b}_{2}$ orbital (belonging primarily to the oxygen), assuming the convention that water is on the $x z$ plane.

The oxygen then holds the formal charge of $+ 1$.